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TRANSCRIPT
135.03-1
Electrical Systerrs - Course 135
FURTHER EXAMPLES OF ELE:TRICAL PROTECTIVE RELAYS
--- ----_._-------------------
]. ",J ~:.NTRODUCTION
Lesson 235.02-1 examined electrical protectiverelays used for motor protection and lesson 235.02-2covere'l the bas ics of protective relays and schemesused to protect busbars , transformers, genera tors andpower lines. This lesson examines, in greater detail,the principles of relays used to protect generators,transformers etc. The next two lessons will examinehow these relays are connected to form composite acand dc protective schemes for transformers, generatorsand other items of electrical equipment in a generating station or heavy water plant.
2.0 DIFFERENTIAL RELAYS
2.1 principle of Operation
Figure 1 (a) shows the princ iple of d ifferential protection applied to a healthy busbarcircuit. Current transformers (CT's) of equalratio are installed on the external side of thebreakers which protect a section of bus bar Bl' Acurrent operated relay is connected between twowires interconnecting the CT's. During healthyconditions, equal currents will flow in and outof the circuit and because the CT's have equalratios, each CT will give an equal output. TheCT outputs 151 and 152 will circulate aroundthe interconnecting circuit. Note, that thecurrents 1n the relay, flow in opposite directions and therefore cancel.
directhe dc
Thisin the
shows the instantaneousflow and Figure l(b)
ac secondary c ircui t.the current behaviour
Figure lea)tions of currentequivalent of thecircuit illustratessecondary circuit.
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••
135.03-1
I
II
I IL J
•
I,-L2
1R=0I S2
DIFfERENTIAl RElAY
Figure l(a): Healthy busbar - currents in relay balance;relay 87-Bl does not operate.
+ ----,..-- 6 V 87 6V ---'- +
Figure l(b): DC equivalent of the ac sc'condary circuit.
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In Figure 2(a) the primary currents 11 and12 are not equal and consequently the secondarycurrents are not equal. There will be a resul tant current in the relay causing it to operateand trip the two breakers 52-Ll and 52-L2, deenergizing the busbar. Figure 2(b) shows the dcequivalent of the ~s secondary circuit.
Because the differential relay (87-B1) willnot operate with load current or "through faults"ie, faul ts outs ide the protected zone, it can beset to operate at a low value of current therebyg 'ling rapid operation when a fault occurs. Atypical setting for relay 87-Bl is 20% of fullload current. There is no need to time delay theoperation of relay 87-Bl and therefore a fastacting attracted armature type of relay is used.
L2L1
PROTECTED ZONEr --- - ---- ClflCWT-BREAKE;S- - - - ---~I EJB OPERATED BY ~ II 52 L1 ""-D1FFERENTIAL RELAY ~ 52 l2 I
I'I, I I ~-+- , 1 _
•••IIi'~••'. BUSBAR ...:~.~~••_
I 1I 1I FAULT I
1 I
: If =1,-I; :L ..J
lsi
Figure 2(a): Unhealthy or faulted busbar - relay operateswith current I s 1 - I s 2
+
A Av v
tt--~8V 87 4V -r-
+
Figure 2(b): DC equivalent of the ac secondary circuit.
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Protection of "T" circuits. Figure 3 showsa "T" circuit protected by differential protection. For sirrplicity, the current is shown toenter via 1 ine 1 and leave in equal amounts vialines 2 and 3. Note:
(a) the protected zone.
(b) the flow of currents in the CT's and secondary circuits.
(c) there is no current flowing in the relay.
If a fault now occurs inside the zone, thecurrents entering and leaving the zone will notbalance. Current will flow in the relay whichwill be connected to trip all the three breakers.This type of protection is normally used where"T" connections are errployed on bus bars andgenerator circuits.
L2
QARYT
PRIMARY---------------------------------------~PROTECTED ZONE Ir-- - ,CURRENT
UPPLY,
FLOW,~ & & 0 0 I . . -~.~ • .v
".v ~
L1 ..... .......... '---- '--
, SECONCURRENFLOW
I 0 IL____________________ ..------------------
~-, .( .. C:}rA< a
87
•L3
s
LOAD
Figu re 3: Diferent ial protec tion appl·ied to a "T" feeder.Note primary and secondary current flow.
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135.03-1
2.2 Types of Differential Relay
2.2.1 Simple differential relays. The differential relays shown in Figures l(a), 2(a)and 3 are simple current operatedattracted-ecm",ture relays. Although thisbasic type 0f relay can be used for protecting gen"ra tors and transformers., theyare rarely used because:
(a) The CT's feeding the relays, when anout-of-zone fault occurs, may notgive exac tly balanced outputs. Anyunbalance in the output from the CT'smay cause the relay to malfunction.The example given in Figure 4 showsthe s i tua tion which can occur whentwo CT' s each of 500/5 ratio, andeach having a 1% error, are subjectedto a through or out of zone faultcurrent of 10 kA. The relay, set at20% of SA or II'., will have a currentof 21'. flowing in it and it will maloperate causing the breakers to tripincorrectly for the out of zonefaul t.
"'"!" ....j BUSBAR OR~, GENERATOR
...... WINDING......
" " " "'lOlA
CT500/5...
FAULTCURRENT
10kA
SECONDARY CURRENT~99'?o X 10000 X5 ~99A--sao- RESULTANT CURRENT
IN RELAY=101A- 99A = 2A
SECONDARY CURRENT=101% X10000 X5=101A
500
Figure 4: Current transformer errors causing differentialrelay to mal-operate on an out of zone fault.
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(b) In the primary circuit, there may betransformers with tap changers. Tapchangers cause the transformer toproduce a different voltage and current ratio. Unless the CT's alsohave taps, (which is very rarely thecase), when the transformer is on atap other than the one giving thenominal ratio there will be >an unbalance in the CT output currents.
2.2.2 Percentage Differential Relays. To overcome these problems detailed in 2.2.1 (a)and (b), relays are made with restraintcoils as well as operating coils. Theserestraint coils prevent tripping duringthrough a "out-oE-zone" faul t cond i tions,but should an "in-zone" fault occur,tripping will take place.
500/5...•....~~~ -1 BUSBAR OR
GENERATORWINDING
500/5--
99A
OPERATECOIL,
100 TURNS
---RESTRAINT COILS10 TURNS EACH
101A
FAULTCURRENT;10 kA
Figure 5(a): Differential relay with restraint coils.Relay does not mal-function when an outof zone Eault occurs.
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135.03-1
Figure 5(a) shows how the restraint andoperating coils are connected. A typicalrelay is considered, which has 100 turnson its operate coil and 10 turns on eachof its restraint coils. Under the samefault cond +ions as shown in Figure 4, therestrain~:oils produce (99 x 10) + (101 x10) = 2000 ampere-turns. The operate coilproduces 2 x 100 = 200 ampere" turns.Clearly, the restraint coils produce agreater magnetic force and the relay doesnot operate. For the relay to operate,the ampere-turns of the operating coilJr.'.•. " t exceed the ampere-turns of there3training coil.
Figure 5(b) shows how a balanced beam typeof differential relay is used. Underhealthy or out of zone fault conditions,the restraint coils have a greater pulland prevent tripping. When an inzonefault occurs, the operate coil has agrea ter pull and the relay operatestripping the appropriate breakers. Thebreakers are not shown in Figure 5(b).
CTCT- ~ BUSBAR OR ~_ SODA
~ v GENERATORWINDING v ~
~,• .. T -5A
BALANCED BEAM G •,SA~AV~·TRIP •
---PIVOr
RESTRAINT/i fl"COILS
LOPERAT;COIL
Figure 5( b): Balanced beam type differential relay.
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Most manufacturers provide tappingsrestraint coils. This enables theor percentage of the restraintaltered to allow for:
on theamountto be
(al CT ratio errors.
(bl the unbalance in CT outputs producedby tap changers on transformers., Seesection 2.2.1(bl.
To overcome these problems, sufficient(but not too much) restraJ.nt has to beprovided and this is done by selecting thecorrect tap. Figure 6 shows, for variouspercentage tap settings of the restraintcoils, curves of through or out of zonefault current versus levels of current tooperate the relay.
20 40 60 80 100 120 140 160 180 200CURRENT IN RESTRAINT COILS AMPS
15% RESTRAINTSETTING
20% RESTRAINTSETTING
RELAYDOES NOTOPERATE
REGION
30 %RESTRAINTSETTING
RELAYOPERATEREGION
30 -----------
25
5
AMPS
,AMINIMUMOPERATINGCURRENT
Figure 6: Typical operating chara~teristics of apercentage differential relay.
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'faking the example shown in Figure 4, ifthe relay is set at 20%, see Figure 6, and100 A flows in each of the restraint coilseach having 10 turns, 100 x 10 x 2 = 2000ampere turns of restraint force are produced.
Before the relay can operate, a Eurrentgreater than 20 A must flow through the100 turn operating coil producing morethan 20 x 100 = 2000 ampere turns ofoperating force. See point A in Figure 6on 20% restraint setting, (also called 20%Slr)2) •
2.2.3 Harmonic Restraint Differential Relays.When transformers are switched onto theirsupply, there is a surge or inrush ofcurrent taken from the supply. Consequently, only the CT' s on the supplysic'le give an output which will cause asimple or a percentage differential relayto operate. It is worth noting that thesurge only lasts for 10 - 20 cycles.
To overcome this problem, relays are producec'l which have a time delay which prevents tripping for approximately 0.5 sec.Unfortunately, the longer the tripping isdelayec'l, the greater the damage that canbe produced in the transformer. Analysisof the surge or inrush current shows thatthe current consists of appreciablequantities of second and third harmonicsof 60 Hz, ie, 120 and 180 Hz. Filters aretherefore provided which ensure that:
(a) only 60 Hz currents flow through theoperate coil.
(bl the 120 and 180 Hz currents are usedto produce additional restraint forthe relay.
Figure 7 shows how a balanced beam relaycan be used to provide the operate,restraint and harmonic restraint featuresthat are required. During normal operation, the restraint. coils prevent relayoperation. During a healthy "switch in",when the supply sic'le CT's are only givingan output, restraint is provided by theharmonic restraint coil. At any time,
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CT250ls-.- LOAO SIDE
250A
TRANSFORMER14kV /28kVSUPPLY SIDE
CT500/S-SA SA
AUXILlARY~CURRENT
TRANSFORMER
~1:-+-_.lPERCENTAGETAPS
vzi~2i?2/t'?2~'PPIVOT
~,
60Hz PASSFILTER
RESTRAINT OPERATECOIL COIL
HARMONICRESTRAINTCOIL
Figure 7: Percentage differential relay with harmonicrestraint. Current flow are shown for ahealthy condition.
should an in zone faul t occu r, the operatecoil will provide sufficient operatingforce to tip the beam and send thenecessary trip signals. Note, no timedelay is required in this relay scheme.
The type of relay shown in Figure 7 showsthe essential features for percentagedifferential protection with harmonicrestraint. Some manufacturers, in placeof the beam, use a disc, and others anattracted armature relay.
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135.03-1
In Figure 7 note that both current transformers have the same secondary current of5.0 A. ~'his ensures that under healthyconditions no current will flow in thedifferential relay. This subject iscovered in ~ore detail in the next lesson.
300 OTHER TYPES OF TRANSFORMER PROTECTIVE DEVICE
3.1 Gas and oil Relays
While the differential relay is the mainmeans of ~,otection for most transformers in thelarge sizes and at higher voltages, other measures may be used either alone or in addition tothe differential relay. One such scheme which isparticularly applicable to the oil filled transformer is the gas relay illustrated in Figure 8.
PRESSUREEQUALIZINGSILICONERUeBERDIAPHRAGMAND SEAL
____ CONNECTOR
-AIR VENT
ALARMMICRO SWITCH
PRESSURE EQUALIZING
_---~~~ft--ORf'CE
BElLOWSS..1 oul oil
Comprene••Ir toflax diaphr.;m
TOP OF
~....~....+..................~~~~~:..-.,....:.'<""..,... ....""<'"""""<'"~-.:":~~:",,,l:'""""--- ~:~~SFOR"'ER
BlEEO£R VALVEta r.te•••accumulatedg....
Figure 8: Sectional view of gas and oil relay.
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This relay consists of 2 separate elements,
(a) An upper element which detects the slowaccummulation of gas produced by a faultdeveloping in the transformer. For example,a loose connection will produce arcing inthe oil. The oil will break down producingquanti ties of methane, hyd rogen, acetyleneand other gases. The relay will collectthese gases and when the predetermined levelhas been reached, the float will fall andinitiate an alarm. The big advantage ofthis type of relay is that it detects afault inside the transformer before largefault currents flow causing severe damage.
(b) A lower element operates when a more violenttype of fault occurs in the transformertank. This type of fault produces a pressure surge in the oil which compresses thebellows in the relay. The pressure in thespace between the bellows and the diaphragmalso rises. This makes the diaphragm flexand push the microswitch which initiates atrip of the transformer.
A gas and oil relay must be mounted at thetop of a transformer tank. This will ensure thatit will collect all the gases tla t are produced.
4.0 THERMAL SENSING AND MEASURING DEVICeS
4.1 Direct Measurement
In many instances, it is possible to measurethe temperature of electrical 8quipment directly.An example of this, is the method used for measuring the temperature of the core laminations ina generator or motor. A direct measurement ispossible because the core is grounded. Anotherexample is where the temperature of water cooledwind ings in a large turbo-gene l:-a tor are measu redby measuring the temperature of the cooling waterleaving the windings.
4.2 Indirect Measurement
Measuring the temperature of live conductorsis much more d iff icul t and involves the use ofinsulated sensors or thermal image devices.
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Insulated sensors are occas ionally used tomeasure the winding temperature of small domesticmotors. Because of possible safety implications,they are not used for winding temperature measurement of 3 phase motors in generatingstations.
An indication - but not an accurate me~sure
ment of the winding temperature in a motor can beachieved by measuring the cooling air temperature. A factor is then added on to allow for thetEmperature difference between the conductor andthe cooling air.
Temperature measurement on motors usingthermal relays producing thermal images isexplained in lesson see 235.02-1. This method issatisfactory for use in motors but with oil filled transformers, a more accurate method is used •
OIL FILLEDTRANSFORMER
.......SUSHING
><:::::l4-tWINOINGS
CURRENTTRANSFORMERHEATERS
SENSINGBULB
Figure 9: Principle of transformer winding temperaturemeasuring device.
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135.03-1
Figure 9 shows the principle of transformerwinding temperature measuring device. Under load,the windings and core in a transformer will produceheat which is transferred to the oil. Due to convection, the hottest part of the oil is at the topof the tank. A temperature sensing device situatedat the top of the transformer would record the topoil temperature but not the winding temp.erature.Obviously, the wind ings will operate at a temperature greater than the top oil temperature. To makethe temperature indicator sense a temperature whichis the same as the winding temperature, a smallheater surrounds the sensing bulb which is placedin the hot oil at the top of the tank. This heateris connected to a current transformer whichsuppl ies the heater wi th a fraction of the mainwinding current. It whould now be appreciatedthat:
(a) The winding I 2R and iron losses will heat theoil and due to convection, the hottest oilwill be at the top of the tank. Because thewinding is a source of heat, the hottest partof the wind ing - usually called the wind inghot spot will be hotter than the top oil. Thewinding hot spot temperature will therefore bethe top oil temperature plus an increase dueto the I 2R prod uced in the wind ing.
(b) The winding temperature indicator will indicate the hot spot temperature, ie, the topoil temperature plus an increase due to I2 Rproduced by the heater.
(c) The wind ing tempera tu re ind icator, afterbeing correctly set up by the manufacturer,will give a true thermal image of winding hotspot temperature under all operating conditions.
4.3 Temperature Measurement Using.Change c~~ Resistance
Because of reactance being present, resistance in ac circuits cannot be measured by applying ac voltage and current. However, on dc circuits, temperature is frequently measured byobserving the change in resistance of a coil or awinding and converting this change in resistanceinto a temperature read ing .. This type of measurement is used to measure the temperature of the derotor windings in an ac generator.
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Fi9ure 10 shews how the resistance of coppervaries linearly WI ".h temperature. If the temperature of a windin9 is reduced, then at some temperature -T, the value of resistance will become zero,. e c materia i Lwcc ';e,'; superconducting. Fi9ure 10ilL_-,st teE; this
The cold cor,ltions are represented by thesolid triangle and the hot conditions are represented by the dotted triange. As the triangles are,dmilar, the'1 the magnitudes of their sides are in
:.2ct ratio.
D
ResistCince
Hot or Working ReSistonce ~__~--~-"'~I>"'-T
Cold Resistonce ~---+I IC I
'" J I__ _ _ _ _ _ _ _ __ ..J'---'-----l~~
I I I I temp ·CI: AI o••~
Figure 10: Resistance as a function of temperature.
Therefore, by comparing lengths of the sidesof the triangles,
AB =
CD
Note:
Length A isLength B ISLength C isLength D is
-T plus the cold temperature-T plus the hot or working temperaturethe cold res istancethe hot resistance
The best way of illustrating the above is toconsider an example.
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Example:
A generator rotor takes 4000 A at 400 V at20°C and 4000 A at 480 Volts when operating at itsoperating temperature. The rotor is wound using acopper conductor. What is the operatingtemperature of the conductor?
V 400Cold resistance = I = 4000 = 0.1
Hot resistance 480= 4000 = .12
Substituting the above in the equation:
A e=B D
gives 20 + 234.5B
.1.12
The value of -Tof -234.5°e isobtained fromTable 1.
-T
Length B = .12 x 254.5 = 305°C above -Toe~
Hot Temp = 305° - 234.5° = 70.5°e above ooe
Resistance
Hot Resistance .12..IL - - -;/ i -1Cold Resistance .1..IL /T--t
: 0=.12...n...
C=.l.n. I
~~ J I I
-T J-~----J.~
'
temp act--- A=20a + 234.50
.--_
• 20aC1-----B=30s,Cl.----------I
t----234.so.a -------;...!-- 70.5 .0.°----...;....1(Capper) OaC
Figure 11: Diag ram 111 ust rat ing Example.
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Table 1: Values of -T for Some Common ElectricalConductor Materials
! CONDlICTO; 1~7 c."ERIIIL--------
SilverCopperAluminum'I'u ngs tenNickel
-T (in °C)
-243-234.5-236-202-147
Where a spot or a single reading is required,the voltage and current are measured with the windlng cold and then a further read ing is taken ofvol tage and cu rrent when the wind ing is hot. Acalculation will then give the winding temperature.
BRUSHES
SLiPRINGS
ROTORWINDINGS
~ VOLTAGE
INPUT
CURRENTTRANSDUCER
CURRENT~ VINPUT r-1':-+-I--'-+-,
FIElDj(CIRCUIT~ BREAKER
~~-06 I e--.....+-----'IIII..c---
0+---0
EXCITATiONSYSTEM
DCOUTPUT
COMPONENT BOXGIVES V/l
RATIO
ROTOR TEMPERATUREINDICATOR,CALIBRATED
IN DEGREES
Figure 12: Circuit for rotor temperature indicator.
Where continuous monitoring of temperature isrequired, for example on the dc rotor of a turbinegenerator, an instrument called a rotor temperatureindicator has been devised. This indicator measures the resistance by comparing the ratio ofvoltage to current. Figure 12 shows the basic circuitry of a rotor temperature indicator for aturbo-generator. The indicator is supplied withalarm contacts which warn the operator when therotor temperature limit has been exceeded.
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ASSIGNMENT
1. By considering "in zone" and "through" faults, explain,using labelled diagrams, how differential protection isused to protect a busbar.
2. (a) By considering "in zone" and "through" faults,explain, using labelled diagrams, how differentialprotection is used to protect a bus bar with a Teedconnec t io n.
(b) Will all the current transformers have the sameratio? Why/why not?
3. Explain, using labelled diagrams, why:
(a) percentage differential relays are required forprotecting generators.
(b) percentage differential relays with harmonicrestraint are required for protecting transformers.
4. (a) Explain how a transformer can be protected using agas and oil relay.
(b) State why the gas andinternal components inprotect the rest of the
oil relay cana transformer
circuit.
protect thebut cannot
(c) Given an outline diagram of a gas and oil relay,label the principal components and explain theirfunction.
5. Using a labelled diagram,measure the temperature oftransformer.
explain how to accuratelythe windings of a loaded
6. Using a labelled diagram, explain how tomeasure the temperature of the rotor windingsgenerator. The turbo-generator is assumed toload.
J.R.C. Cowling
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accuratelyin a turbobe on full